Excluded as a possibility, therefore the final probability of the woman having either aa or Aa genotypes is 1/3 and 2/3, respectively

Albinism is a somatic recessive condition resulting from the inability to produce the dark pigment melanin in skin and hair. A man and woman with normal skin pigmentation have two children. The man has one albino parent; the woman has parents with normal pigmentation, but an albino brother.

Case 1: If the woman is Aa, the probability of at least one of the two children being albino from the cross (Aa × Aa) equals the probability that one child is albino (1/4) × the probability that the other child is normal (3/4) × 2 (must multiply by 2 because either birth order is possible for this outcome), plus the probability that both children are albino, (1/4) × (1/4) = 1/16. Therefore, the final probability that at least one child is albino equals 2(3/4)( 1/4) + (1/16) = 6/16 + 1/16 = 7/16 (43.75%).
Case 2: If the woman is AA, the probability of at least one of the two children being albino from the cross (Aa × AA) is zero.

As noted above, if Case 1 is true, then the probability of both children being albino (aa) equals (1/4) × (1/4) = 1/16 (6.25%).